Answer

Regional vs. National Housing Price Comparison Report

Introduction

Purpose

To determine if my region’s housing prices and housing square footage are significantly different from those of the national market. The sample will be from at least one state in the pacific region. The hypothesis to determine is if the prices in the Pacific region market are higher than the national market average and if the square footage for home in the pacific region different from the average square footage for the homes in the national market. The average listing prices for national market is 288407 and the average for square footage for homes in the national market is 142.

Sample

The sample will comprise of 100 observations from at least one state in the Pacific region. The data will be of house listing prices and cost per square foot.

Questions and Type of Test

The first hypothesis is to determine if the house listing prices in the Pacific region is higher than those of the national markets. This hypothesis is directional hence it is a 1-tail test. We will use a t-test in analysis at a confidence interval of 95%.

H₀: µ >288407

H_A: µ ≤ 288407

The first hypothesis is to determine if the square footage for homes in the Pacific region are different than those of the national markets. This hypothesis is non-directional hence it is a 2-tail test. We will use a t-test in analysis at a confidence interval of 95%.

H₀: µ = 142

H_A: µ ≠ 142

1-Tail Test

Hypothesis

The population parameter comprises the entire home sales from the Pacific region.

H₀: µ > 288407

H_A: µ ≤ 288407

The significance level (α) is set at 0.05

Data analysis

=QUARTILE ([data range], [quartile number])

Q1 = 303612.5

Q3 = 585781.1

From the descriptive table above, the sample mean is 485924.7064, the median of the data is 417550, and the standard deviation is 246108.1628. From the graph it is evident that the shape is skewed to the right.

The assumptions for a t-test include the data follows a continuous scale, the data was randomly collected and is a representation of the population, from the histogram the data results in a normal distribution, the sample size is large, and there is homogeneity of variance. The data provided meets all the assumptions.

Hypothesis Test Calculations:

t = (mean – target)/standard error.

The mean is your regional mean(Pacific), and the target is the national mean.

t = (485924.7064 – 288407) / 24610.81628

t = 8.026

Calculate the probability (p value)

=T.DIST.RT([test statistic], [degree of freedom]).

The degree of freedom is calculated by subtracting 1 from your sample size = (100 – 1).

= T.DIST.RT([8.026], [99]).

= 1.056E-12

Interpretation

The p-value (1.056E-12) is less than the significance level (0.05).

Since 1.056E-12 < 0.05: reject the null hypothesis (H₀)

After rejecting the null hypothesis, we conclude that the average of the house listing prices in the Pacific region is not higher than those of the national markets.

2-Tail Test

Hypotheses:

The population parameter comprises the entire home sales from the Pacific region.

H₀: µ = 142

H_A: µ ≠ 142

The significance level (α) is set at 0.05

Data Analysis:

=QUARTILE([data range], [quartile number]) ]

Q1 = 167.3693

Q3 = 347. 4761

From the descriptive analysis, the sample mean is 279.3433, the median is 211.9862, and the standard deviation is 160.486. From the graph it is evident that the data is distributed normal and has a bell shape. The data is skewed to the right.

The assumptions for a t-test include the data follows a continuous scale, the data was randomly collected and is a representation of the population, from the histogram the data results in a normal distribution, the sample size is large, and there is homogeneity of variance. The data provided meets all the assumptions.

Hypothesis Test Calculations:

Determining the test statistic (t).

t = (mean – target)/standard error.

In this case, the mean is the Pacific mean (279.343), and the target is the national mean (142)

t = (279.343 – 142) / 16.05

t = 8.557

Calculating the probability (p value).

=T.DIST.RT([test statistic], [degree of freedom]).

The degree of freedom is calculated by subtracting 1 from your sample size = (100 – 1)

= T.DIST.2T([8.557], [99]).

P–value = 1.516E-13

Interpretation

The p-value (1.516E-13) is less that the significance level (0.05)

Since the 1.516E-13< 0.05: reject the null hypothesis (H₀)

After rejecting the null hypothesis, we conclude that the average of the square footage for homes in the Pacific region are different than those of the national markets

Comparison of the Test Results:

Confidence interval for Pacific region house listing prices

Confidence Interval = sample mean ± margin of error

Margin of error = (z-score * (population standard deviation / square root of sample size))

Z-score at 95% = 1.96

= 1.96 * (163.986 / 10)

= 32.141

CI = 485924.7064 ± 32.141

= (485892.6, 485956.8)

I am 95 % confident that the true mean of the Pacific house listing prices is between 485892.6 dollars and 485956.8 dollars.

Confidence interval for Pacific cost per square foot houses

Margin of error = 1.96 * (92/10)

= 18.032

CI = 279.3433 ± 18.032

= (261.3113, 297.3753)

I am 95 % confident that the true mean of the Pacific cost per square foot of houses is between 261.3113 dollars and 297.3753 dollars.

Final Conclusions

From the sample collected we are able to determine that the houses listing prices in the Pacific is not more than the house listing prices for the nation market.

We have also determined that the cost per square foot of houses in the Pacific region are different from the cost per square foot of national markets.

We have determined the range of values for two samples.

From these statements, we have successfully answered the regional sales director questions.

I was not surprised by the findings because we were using a sample to infer to the population.